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Simplest miniversal deformations of matrices,

matrix pencils, and contragredient matrix

pencils

Ma¯ Isabel Garcıa-Planas

Dept. de Matematica Aplicada IUniversitat Politecnica de Catalunya

Marques de Sentmenat 63-4-3, 08029 Barcelona, Spainigarcia@ma1.upc.es

Vladimir V. Sergeichuk∗

Institute of Mathematics, Tereshchenkivska St. 3Kiev, Ukraine

sergeich@imath.kiev.ua

Abstract

For a family of linear operators A(~λ) : U → U over C that smoothlydepend on parameters ~λ = (λ1, . . . , λk), V. I. Arnold obtained the sim-plest normal form of their matrices relative to a smoothly dependingon ~λ change of a basis in U . We solve the same problem for a familyof linear operators A(~λ) : U → U over R, for a family of pairs of linearmappings A(~λ) : U → V, B(~λ) : U → V over C and R, and for a fam-ily of pairs of counter linear mappings A(~λ) : U → V, B(~λ) : V → U

over C and R.

This is the authors’ version of a work that was published in Linear Algebra Appl.302–303 (1999) 45–61.

∗Partially supported by Grant No. UM1-314 of the U.S. Civilian Research and Devel-opment Foundation for the Independent States of the Former Soviet Union.

1

1 Introduction

All matrices and representations are considered over a field F ∈ {C, R}.We base on ideas and methods from Arnold’s article [1], extending them onquiver representations.

Systems of linear mappings are conveniently studied if we consider themas representations of a quiver. A quiver is a directed graph, its representationA over F is given by assigning to each vertex i a finite dimensional vectorspace Ai over F and to each arrow α : i → j a linear mapping Aα : Ai → Aj .For example, the problems of classifying representations of the quivers

• • •q

1i• •q

i

are the problems of classifying, respectively, linear operators A : U → U(its solution is the Jordan normal form), pairs of linear mappings A : U →V, B : U → V (the matrix pencil problem, solved by Kronecker), and pairsof counter linear mappings A : U → V, B : V → U (the contagredient matrixpencil problem, solved in [2] and studied in detail in [3]).

Studying families of quiver representations smoothly depending on pa-rameters, we can independently reduce each representation to canonical form,but then we lose the smoothness (and even the continuity) relative to the pa-rameters. It leads to the problem of reducing to normal form by a smoothlydepending on parameters change of bases not only the matrices of a givenrepresentation, but of an arbitrary family of representations close to it. Thisnormal form is obtained from the normal form of matrices of the given rep-resentation by adding to some of their entries holomorphic functions of theparameters that are zero for the zero value of parameters. The number ofthese entries must be minimal to obtain the simplest normal form.

This problem for representations of the quiver ·i over C was solved byArnold [1] (see also [4, § 30]). We solve it for holomorphically depending on

parameters representations of the quiver ·i over R and representations of

the quivers · ·q

1 and · ·q

i both over C and over R. In the obtained simplestnormal forms, all the summands to entries are independent parameters. Anormal form with the minimal number of independent parameters, but notof the summands to entries, was obtained in [5] (see also [4, § 30E]) for

representations of the quiver ·i over R and in [6] (partial cases were

considered in [7]–[8]) for representations of the quiver · ·q

1 over C.

2

2 Deformations of quiver representations

Let Q be a quiver with vertices 1, . . . , t. Its matrix representation A ofdimension ~n = (n1, . . . , nt) ∈ {0, 1, 2, . . .}t over F is given by assigning amatrix Aα ∈ Fnj×ni to each arrow α : i → j. Denote by R(~n, F) the vectorspace of all matrix representations of dimension ~n over F. An isomorphismS : A → B of A, B ∈ R(~n, F) is given by a sequence S = (S1, . . . , St) ofnon-singular matrices Si ∈ Gl(ni, F) such that Bα = SjAαS−1

i for each arrowα : i → j.

By an F-deformation of A ∈ R(~n, F) is meant a parametric matrix rep-

resentation A(λ1, . . . , λk) (or for short A(~λ), where ~λ = (λ1, . . . , λk)), whoseentries are convergent in a neighborhood of ~0 power series of variables (theyare called parameters) λ1, . . . , λk over F such that A(~0) = A.

Two deformations A(~λ) and B(~λ) of A ∈ R(~n, F) are called equivalent

if there exists a deformation I(~λ) (its entries are convergent in a neigh-borhood of ~0 power series and I(~0) = I) of the identity isomorphism I =(In1

, . . . , Int) : A → A such that

Bα(~λ) = Ij(~λ)Aα(~λ)I−1i (~λ), α : i → j,

in a neighborhood of ~0.A deformation A(λ1, . . . , λk) of A is called versal if every deformation

B(µ1, . . . , µl) of A is equivalent to a deformation A(ϕ1(~µ), . . . , ϕk(~µ)), whereϕi(~µ) are convergent in a neighborhood of ~0 power series such that ϕi(~0) = ~0.A versal deformation A(λ1, . . . , λk) of A is called miniversal if there is noversal deformation having less than k parameters.

For a matrix representation A ∈ R(~n, F) and a sequence C = (C1, . . . , Ct),Ci ∈ F

ni×ni , we define the matrix representation [C, A] ∈ R(~n, F) as follows:

[C, A]α = CjAα − AαCi, α : i → j.

A miniversal deformation A(λ1, . . . , λk) of A will be called simplest if it isobtained from A by adding to certain k of its entries, respectively, λ1 to thefirst, λ2 to the second, . . . , and λk to the kth. The next theorem is a simpleconclusion of a well known fact.

Theorem 2.1. Let A(~λ) = A+B(~λ), ~λ = (λ1, . . . , λk), be an F-deformation

of a matrix representation A ∈ R(~n, F), F ∈ {C, R}, where k entries of B(~λ)

3

are the independent parameters λ1, . . . , λk and the other entries are zeros.Then A(~λ) is a simplest miniversal deformation of A if and only if

R(~n, F) = PA ⊕ TA,

where PA is the k-dimensional vector space of all B(~a), ~a ∈ F k, and TA isthe vector space of all [C, A], C ∈ Fn1×n1 × · · · × Fnt×nt.

Proof. Two subspaces of a vector space V are transversal if their sum is equalto V . The class of all isomorphic to A ∈ R(~n, F) matrix representations maybe considered as the orbit AG of A under the following action of the groupG = GL(n1, F) × · · · × GL(nt, F) on the space R(~n, F):

ASλ = SjAλS

−1i , λ : i → j,

for all A ∈ R(~n, F), S = (S1, . . . , St) ∈ G, and arrows λ. A deformation

A(~λ) of a matrix representation A ∈ R(~n, F) is called a transversal to theorbit AG at the point A if the space R(~n, F) is the sum of the space A∗F

k

(that is, of the image of the linearization A∗ of A(~λ) near A; the linearizationmeans that only first derivatives matter) and of the tangent space to the orbitAG at the point A. The following fact is well known (see, for example, [9,Section 1.6] and [1]): a transversal (of the minimal dimension) to the orbitis a (mini)versal deformation.

It proves the theorem since PA is the space A∗Fk and TA is the tangent

space to the orbit AG at the point A; the last follows from

AI+εCλ = (I + εCj)Aλ(I + εCi)

−1 = (I + εCj)Aλ(I − εCi + ε2Ci − · · · )

= Aλ + ε(CjAλ − AλCi) + ε2... ,

for all C = (C1, . . . , Ct), Ci ∈ Fni×ni, small ε, and arrows λ : i → j.

Corollary 2.1. There exists a simplest miniversal F-deformation for everymatrix representation over F ∈ {C, R}.

Proof. Let A ∈ R(~n, F), let T1, . . . , Tr be a basis of the space TA, and letE1, . . . , El be the basis of R(~n, F) consisting of all matrix representations ofdimension ~n such that each of theirs has one entry equaling 1 and the othersequaling 0. Removing from the sequence T1, . . . , Tr, E1, . . . , El every repre-sentation that is a linear combination of the preceding representations, we

4

obtain a new basis T1, . . . , Tr, Ei1 , . . . , Eik of the space R(~n, F). By Theorem2.1, the deformation

A(λ1, . . . , λk) = A + λ1Ei1 + · · · + λkEik

is a simplest miniversal deformation of A since Ei1 , . . . , Eik is a basis of PAand R(~n, F) = PA ⊕ TA.

By a set of canonical representations of a quiver Q, we mean an arbitraryset of “nice” matrix representations such that every class of isomorphic rep-resentations contains exactly one representation from it. Clearly, it sufficesto study deformations of the canonical representations.

Arnold [1] obtained a simplest miniversal deformation of the Jordan ma-

trices (i.e., canonical representations of the quiver ·i ). In the remainingof the article, we obtain simplest miniversal deformations of canonical repre-

sentations of the quiver ·i over R and of the quivers · ·q

1 and · ·q

i bothover C and over R.

Remark 2.1. Arnold [1] proposed an easy method to obtain a miniversal (butnot a simplest miniversal) deformation of a matrix under similarity by solvinga certain system of linear equations. The method is of considerable currentuse (see [6, 7, 8, 10]). Although we do not use it in the next sections, nowwe show how to extend this method to quiver representations.

The space R(~n, F) may be considered as a Euclidean space with scalarproduct

〈A, B〉 =∑

α∈Q1

tr(AαB∗α),

where Q1 is the set of arrows of Q and B∗α is the adjoint of Bα.Let A ∈ R(~n, F) and let T1, . . . , Tk be a basis of the orthogonal comple-

ment T ⊥A to the tangent space TA. The deformation

A(λ1, . . . , λk) = A + λ1T1 + · · ·+ λkTk (1)

is a miniversal deformation (since it is a transversal of the minimal dimensionto the orbit of A) called an orthogonal miniversal deformation.

For every arrow α : i → j, we denote b(α) := i and e(α) := j. Bythe proof of Theorem 2.1, B ∈ T ⊥A if and only if 〈B, [C, A]〉 = 0 for all

5

C ∈ Fn1×n1 × · · · × Fnt×nt . Then

〈B, [C, A]〉 =∑

α∈Q1

tr(Bα(Ce(α)Aα − AαCb(α))∗)

=∑

α∈Q1

tr(BαA∗αC∗e(α) − BαC∗b(α)A∗α) =

t∑

i=1

tr(SiC∗i ) = 0,

whereSi :=

∑

e(α)=i

BαA∗α −∑

b(α)=i

A∗αBα.

Taking Ci = Si for all vertices i = 1, . . . , t, we obtain Si = 0.Therefore, every orthogonal miniversal deformation of A has the form

(1), where T1, . . . , Tk is a fundamental system of solutions of the system ofhomogeneous matrix equations

∑

e(α)=i

XαA∗α =∑

b(α)=i

A∗αXα, i = 1, . . . , t,

with unknowns T = {Xα |α ∈ Q1}.

3 Deformations of matrices

In this section, we obtain a simplest miniversal R-deformation of a real matrixunder similarity.

Let us denote

JC

r (λ) = Jr(λ) :=

λ 1

λ. . .. . . 1

λ

, Jr := Jr(0); (2)

and, for λ = a + bi ∈ C (b > 0), denote JR

r (λ) := Jr(λ) if b = 0 and

JR

r (λ) :=

Tab I2

Tab. . .. . . I2

Tab

if b > 0, where Tab :=

[

a b−b a

]

, (3)

6

(the size of Jr(λ), JC

r (λ) and JR

r (λ) is r × r).Clearly, every square matrix over F ∈ {C, R} is similar to a matrix of the

form⊕i ΦF(λi), λi 6= λj if i 6= j, (4)

uniquely determined up to permutations of summands, where

ΦF(λi) := diag(JF

si1(λi), JF

si2(λi), . . . ), si1 > si2 > · · · . (5)

LetH = [Hij ] (6)

be a parametric block matrix with pi × qj blocks Hij of the form

Hij =

[

∗... 0∗

]

if pi 6 qj , Hij =

[

0∗ · · · ∗

]

if pi > qj , (7)

where the stars denote independent parameters.Arnold [1] (see also [4, § 30]) proved that one of the simplest miniversal

C-deformations of the matrix (4) for F = C is ⊕i(ΦC(λi)+Hi), where Hi is of

the form (6). Galin [5] (see also [4, § 30E]) showed that one of the miniversalR-deformations of the matrix (4) for F = R is ⊕i(Φ

R(λi) + Hλi), where Hλ

(λ ∈ R) is of the form (6) and Hλ (λ /∈ R) is obtained from a matrix of theform (6) by the replacement of its entries α + βi with 2 × 2 blocks Tαβ (see(3)). For example, a real 4 × 4 matrix with two Jordan 2 × 2 blocks witheigenvalues x ± iy (y 6= 0) has a miniversal R-deformation

x y 1 0−y x 0 10 0 x y0 0 −y x

+

α1 β1 0 0−β1 α1 0 0α2 β2 0 0−β2 α2 0 0

(8)

with the parameters α1, β1, α2, β2. We prove that a simplest miniversal R-deformation of this matrix may be obtained by the replacement of the secondcolumn (β1, α1, β2, α2)

T in (8) with (0, 0, 0, 0)T .

Theorem 3.1 (Arnold [1] for F = C). One of the simplest miniversal F-deformations of the canonical matrix (4) under similarity over F ∈ {C, R}is ⊕i(Φ

F(λi) + Hi), where Hi is of the form (6).

7

Proof. Let A be the matrix (4). By Theorem 2.1, we must prove that forevery M ∈ Fm×m there exists S ∈ Fm×m such that

M + SA − AS = N, (9)

where N is obtained from ⊕iHi by replacing its stars with elements of F andis uniquely determined by M . The matrix A is block-diagonal with diagonalblocks of the form JF

r (λ). We apply the same partition into blocks to M andN and rewrite the equality (9) for blocks:

Mij + SijAj − AiSij = Nij .

The theorem follows from the next lemma.

Lemma 3.1. For given JF

p (λ), JF

q (µ), and for every matrix M ∈ F p×q thereexists a matrix S ∈ F p×q such that M + SJF

q (µ) − JF

p (λ)S = 0 if λ 6= µ,and M + SJF

q (µ) − JFp (λ)S = H if λ = µ, where H is of the form (7) with

elements from F instead of the stars; moreover, H is uniquely determined byM .

Proof. If λ 6= µ then JFq (µ) and JF

p (λ) have no common eigenvalues, thematrix S exists by [11, Sect. 8].

Let λ = µ and let F = C or λ ∈ R. Put C := SJF

q (λ) − JF

p (λ)S =SJq − JpS. As is easily seen, C is an arbitrary matrix [cij] (for a suitable S)satisfying the condition: if its diagonal Ct = {cij | i − j = t} contains bothan entry from the first column and an entry from the last row, then the sumof entries of this diagonal is equal to zero. It proves the lemma in this case.

Let λ = µ, F = R and λ = a + bi, b > 0. Then p = 2m and q = 2nfor certain m and n. We must prove that every 2m × 2n matrix M canbe reduced to a uniquely determined matrix H of the form (7) (with realnumbers instead of the stars) by transformations

M 7−→ M + SJR

2n(λ) − JR

2m(λ)S, S ∈ F2m×2n. (10)

Let us partition M and S into 2 × 2 blocks Mij and Sij , where 1 6 i 6 mand 1 6 j 6 n. For every 2 × 2 matrix P = [pij ], define (see (3))

P ′ := PT01 − T01P =

[

−p12 − p21 p11 − p22

p11 − p22 p12 + p21

]

.

8

By (3), the transformation (10) has the form M 7→ M + S(Tab ⊕ · · · ⊕Tab)− (Tab ⊕ · · · ⊕ Tab)S + SJ2

2n − J22mS = M + b[S(T01 ⊕ · · · ⊕ T01)− (T01 ⊕

· · · ⊕ T01)S] + SJ22n − J2

2mS, that is

M 7−→

M11 M12 · · · M1n

M21 M22 · · · M2n

. . . . . . . . . . . . . . . . . . . . . .Mm1 Mm2 · · · Mmn

+ b

S ′11 S ′21 · · · S ′1n

S ′21 S ′22 · · · S ′2n

. . . . . . . . . . . . . . . . . . . .S ′m1 S ′m2 · · · S ′mn

+

0 S11 · · · S1,n−1

0 S21 · · · S2,n−1

. . . . . . . . . . . . . . . . . . . .0 Sm1 · · · Sm,n−1

−

S21 S22 · · · S2n

. . . . . . . . . . . . . . . . . . . .Sm1 Sm2 · · · Smn

0 0 · · · 0

. (11)

Let first m 6 n. If m > 1, we make Mmn = 0 selecting S ′mn and Sm,n−1.To preserve it, we must further take the transformations (11) with S satis-

fying bS ′mn + Sm,n−1 = 0; that is, S ′mn = −b−1Sm,n−1 and Sm,n−1 =[

−α

β

β

α

]

with arbitrary α and β.

Selecting S ′m,n−1 =[

−2β

−2α

−2α

2β

]

and Sm,n−2, we make Mm,n−1 = 0. To pre-

serve it, we must take bS ′m,n−1 + Sm,n−2 = 0; that is, S ′m,n−1 = −b−1Sm,n−2

and Sm,n−2 =[

−α

β

β

α

]

with arbitrary α and β; and so on until obtain Mm2 =

· · · = Mmn = 0. To preserve theirs, we must take Sm1 =[

−α

β

β

α

]

with arbi-

trary α and β and suitable Sm2, . . . , Smn. Then Mm1 7→ Mm1 + b[

−2β

−2α

−2α

2β

]

,

we make Mm1 =[

γ

δ

00

]

, where γ and δ are uniquely determined.We have reduced the last strip of M to the form

[Mm1 · · ·Mmn] =

[

γ 0 · · · 0δ 0 · · · 0

]

. (12)

To preserve it, we must take Sm1 = · · · = Sm,n−1 = S ′mn = 0 since thenumber of zeros in Mm1, . . . , Mmn is equal to the number of parameters inSm1, . . . , Sm,n−1, S

′mn.

The next to last strip of M transforms as follows: [Mm−1,1 · · ·Mm−1,n] 7→[Mm−1,1 · · ·Mm−1,n]+b[S ′m−1,1 · · ·S

′m−1,n]+[0 Sm−1,1 · · ·Sm−1,n−1]−[0 · · ·0 Smn].

In the same way, we reduce it to the form

[Mm−1,1 · · ·Mm−1,n] =

[

τ 0 · · · 0ν 0 · · · 0

]

9

taking, say, Smn = 0. We must prove that τ and ν are uniquely determinedfor all Smn such that S ′mn = 0. It may be proved as for the γ and δ from (12)since the next to last horizontal strip of M , without the last block, is trans-formed as the last strip: [Mm−1,1 · · ·Mm−1,n−1] 7→ [Mm−1,1 · · ·Mm−1,n−1] +b[S ′m−1,1 · · ·S

′m−1,n−1] + [0 Sm−1,1 · · ·Sm−1,n−2] (recall that m 6 n, so this

equality is not empty for m > 1).We repeat this procedure until reduce M to the form (7).If m > n, we reduce M to the form (7) starting with the first vertical

strip.

4 Deformations of matrix pencils

The canonical form problem for pairs of matrices A, B ∈ Fm×n under trans-formations of simultaneous equivalence

(A, B) 7→ (SAR−1, SBR−1), S ∈ GL(m, F), R ∈ GL(n, F),

(that is, for representations of the quiver · ·q

1 ) was solved by Kronecker:each pair is uniquely, up to permutation of summands, reduced to a directsum of pairs of the form (see (2)–(3))

(I, JF

r (λ)), (Jr, I), (Fr, Kr), (F Tr , KT

r ), (13)

where λ = a + bi ∈ C (b > 0 if F = R) and

Fr =

1 0

0. . .. . . 1

0 0

, Kr =

0 0

1. . .. . . 0

0 1

(14)

are matrices of size r × (r − 1), r × (r − 1), r > 1.A miniversal, but not a simplest miniversal, deformation of the canonical

pairs of matrices under simultaneous similarity was obtained in [6], partialcases were considered in [7]–[8].

Denote by 0↑ (resp., 0↓, 0←, 0→) a matrix, in which all entries are zeroexcept for the entries of the first row (resp., the last row, the first column,the last column) that are independent parameters; and denote by Z the p×q

10

matrix, in which the first max{q−p, 0} entries of the first row are independentparameters and the other entries are zeros:

0↑ =

∗ · · · ∗0 · · · 0. . . . . . . . .0 · · · 0

, Z =

∗ · · · ∗ 0 · · · 0. . .

0 0 · · · 0

. (15)

Theorem 4.1. Let

(A, B) =l

⊕

i=1

(Fpi, Kpi

) ⊕ (I, C) ⊕ (D, I) ⊕r

⊕

i=1

(F Tqi

, KTqi) (16)

be a canonical pair of matrices under simultaneous equivalence over F ∈{C, R}, where C is of the form (4), D = ΦF(0) (see (5)), and1 p1 6 . . . 6 pl,q1 > . . . > qr. Then one of the simplest miniversal F-deformations of (A, B)has the form (A,B) =

Fp1

Fp2

. . .Fpl

0

0 0

0

I

D

0

0

F Tq1

F Tq2

. . .F T

qr

0

0

0↓

0↓

...0↓

0→ 0→ · · · 0→

0→· · ·0→0→,

Kp1

Kp2

. . .Kpl

0

0

C

I

0

KTq1

KTq2

. . .KT

qr

0

Z · · · Z...Z

. . .

ZT · · · ZT

...ZT

. . .

0↑

0↑

...0↑

0↑

0↑

...0↑

0← 0← · · · 0←

0

0

,

where C and D are simplest miniversal F-deformations of C and D undersimilarity (for instance, given by Theorem 3.1).

Let us denote by S≻ (resp., S≺, Sg, Sf) the matrix that is obtainedfrom a matrix S by removing of its first column (resp., last column, first row,

1We use a special ordering of summands in the decomposition (16) to obtain A and Bin the upper block triangular form except for blocks in C and D.

11

last row), and denote by S⊲ (resp., S⊳, S▽, S△) the matrix that is obtainedfrom a matrix S by connecting of the zero column to the right (resp., zerocolumn to the left, zero row at the bottom, zero row at the top).

The following equalities hold for every p × q matrix S:

SFq = S≺ SKq = S≻ SF Tq+1 = S⊲ SKT

q+1 = S⊳ SJq = S≺⊳

Fp+1S = S▽ Kp+1S = S△ F Tp S = Sf KT

p S = Sg JpS = Sg

▽

Proof of Theorem 4.1. By Theorem 2.1, we must prove that for every M, N ∈Fm×n there exist S ∈ Fm×m and R ∈ Fn×n such that

(M, N) + (SA − AR, SB − BR) = (P, Q), (17)

where (P, Q) is obtained from (A,B) − (A, B) by replacing the stars withelements of F and is uniquely determined by (M, N). The matrices A and Bhave the block-diagonal form: A = A1⊕A2⊕· · · , B = B1⊕B2 ⊕· · · , wherePi = (Ai, Bi) are direct summands of the form (13). We apply the samepartition into blocks to M and N and rewrite the equality (17) for blocks:

(Mij , Nij) + (SijAj − AiRij , SijBj − BiRij) = (Pij , Qij),

Therefore, for every pair of summands Pi = (Ai, Bi) and Pj = (Aj , Bj),i 6 j, we must prove that

(a) the pair (Mij , Nij) can be reduced to the pair (Pij , Qij) by transfor-mations (Mij , Nij) 7→ (Mij , Nij) + (△Mij ,△Nij), where

△Mij := SAj − AiR, △Nij := SBj − BiR

with arbitrary R and S; moreover, (Pij, Qij) is uniquely determined (more ex-actly, its entries on the places of stars are uniquely determined) by (Mij , Nij);and, if i < j,

(b) the pair (Mji, Nji) can be reduced to the pair (Pji, Qji) by transfor-mations (Mji, Nji) 7→ (Mji, Nji) + (△Mji,△Nji), where

△Mji := SAi − AjR, △Nji := SBi − BjR

with arbitrary R and S; moreover, (Pji, Qji) is uniquely determined by(Mji, Nji).

12

Case 1: Pi = (Fp, Kp) and Pj = (Fq, Kq), p 6 q.

(a) We have △Mij = SFq − FpR = S≺ − R▽. Adding △Mij ,we make Mij = 0; to preserve it, we must further take S and R for

which △Mij = 0, i.e. S = [R▽

... ], where the points denote an arbitrary

column. Further, △Nij = SKq − KpR = S≻ − R△ = [R▽

... ]≻ − R△ =

[X▽

... ] − [...X]△, where X := R≻. Clearly, △Nij is an arbitrary matrix

[δαβ] that satisfies the condition: if its diagonal Dt = {δαβ |α − β = t}contains an entry from the first row and does not contain an entry fromthe last column, then the sum of entries of this diagonal is equal to zero.Adding △Nij, we make Nij = Z, where Z is of the form (15) but withelements of F instead of the stars. If i = j, then p = q, Nii = Z hassize p × (p − 1), so Nii = 0 (see (15)).

(b) We have △Mji = SFp − FqR and △Nji = SKp − KqR; so weanalogously make Mji = 0 and Nji = Z. But since Z has size q×(p−1)and p 6 q, Nji = Z = 0 (see (15)).

Case 2: Pi = (Fp, Kp) and Pj = (I, JFq (λ)).

(a) We have △Mij = S−FpR = S−R▽. Make Mij = 0; to preserveit, we must further take S = R▽. Then △Nij = SJF

q (λ) − KpR =(RJF

q (λ))▽ −R△. Using the last row of R, we make the last row of Nij

equaling zero, then the next to the last row equaling zero, and so onutil reduce Nij to the form 0↑ (with elements of F instead of the stars).

(b) We have △Mji = SFp − R = S≺ − R. Make △Mji = 0, thenR = S≺; △Nji = SKp − JF

q (λ)R = S≻ − (JFq (λ)S)≺. We make Nji = 0

starting with the last row (with the last horizontal strip if F = R andλ /∈ R).

Case 3: Pi = (Fp, Kp) and Pj = (Jq, I).

(a) We have △Nij = S−KpR, make Nij = 0, then S = KpR = R△;△Mij = SJq −FpR = (RJq)△−R▽. Reduce Mij to the form 0↓ startingwith the first row.

(b) We have △Nji = SKp −R, make △Nji = 0, then R = SKp =S≻; △Mji = SFp − JqR = S≺ − (JqS)≻. We make Mji = 0 startingwith the last row.

Case 4: Pi = (Fp, Kp) and Pj = (F Tq , KT

q ).

13

(a) We have △Mij = SF Tq − FpR = S⊲ −R▽. Reduce Mij to the

form 0→, then (S⊲ − R▽)≺ = S − R≺▽

= 0, S = R≺▽. Put X := R≺,

then S = X▽ and R = [X... ], where the points denote an arbitrary

row. Further, △Nij = SKTq − KpR = S⊳ − R△ = (X▽)⊳ − [X

... ]△.Clearly, △Nij is an arbitrary matrix [δαβ ] that satisfies the condition:if its secondary diagonal Dt = {δαβ |α + β = t} contains an entry fromthe first row, then the sum of entries of this diagonal is equal to zero.Adding △Nij, we reduce Nij to the form 0↑.

(b) We have △Mji = SFp−F Tq R = S≺−Rf. Make Mji = 0, then

S = [Rf... ]. Further, △Nji = SKp −KT

q R = S≻−Rg = [Rf... ]≻−Rg,

make Nji = 0 starting with the last column.

Case 5: Pi = (I, JF

p (λ)) and Pj = (I, JF

q (µ)).

(a) We have △Mij = S − R. Make Mij = 0, then S = R;△Nij = SJF

q (µ) − JFp (λ)R. Using Lemma 3.1, we make Nij = 0 if

λ 6= µ and Nij = H if λ = µ.

(b) We have △Mji = S − R and △Nji = SJF

p (λ) − JF

q (µ)R. Asin Case 5(a), make Mji = 0, Nji = 0 if λ 6= µ and Nji = H if λ = µ.

Case 6: Pi = (I, JF

p (λ)) and Pj = (Jq, I).

(a) We have △Mij = SJq − R = S≺⊳− R. Make Mij = 0, then

R = S≺⊳

; △Nij = S − JF

p (λ)R = S − (JF

p (λ)S)≺⊳. We make Nij = 0

starting with the first column.

(b) We have △Mji = S − JqR, make Mji = 0, then S = Rg

▽;

△Nji = SJFp (λ)−R = (RJF

p (λ))g

▽−R. We make Nji = 0 starting with

the last row.

Case 7: Pi = (I, JF

p (λ)) and Pj = (F Tq , KT

q ).

(a) We have △Mij = SF Tq − R. Make Mij = 0, then R = S⊲;

△Nij = SKTq − JF

p (λ)R = S⊳− (JF

p (λ)S)⊲. We reduce Nij to the form0← starting with the last row (with the last horizontal strip if F = R

and λ /∈ R).

(b) We have △Mji = S − F Tq R, make Mji = 0, then S = Rf,

△Nji = SJF

p (λ)−KTq R = (RJF

p (λ))f−Rg. We make Nji = 0 startingwith the first column (with the first vertical strip if F = R and λ /∈ R).

14

Case 8: Pi = (Jp, I) and Pj = (Jq, I). Interchanging the matrices in eachpair, we reduce this case to Case 5.

Case 9: Pi = (Jp, I) and Pj = (F Tq , KT

q ).

(a) We have △Nij = SKTq − R. Make Nij = 0, then R = S⊳;

△Mij = SF Tq − JpR = S⊲ − (JpS)⊳. We reduce Mij to the form 0→

starting with the first column.

(b) We have △Nji = S − KTq R, make Nji = 0, then S = Rg,

△Mji = SJp − F Tq R = (RJp)

g − Rf. We make Mji = 0 starting withthe first column.

Case 10: Pi = (F Tp , KT

p ) and Pj = (F Tq , KT

q ), p > q.

(a) We have △Mij = SF Tq − F T

p R and △Nij = SKTq − KT

p R, so(△Mij)

T = (−RT )Fp−Fq(−ST ) and (△Nij)T = (−RT )Kp−Kq(−ST ).

Reasoning as in Case 1(a), we make MTij = 0 and NT

ij = Z, that isMij = 0 and Nij = ZT (Nij = 0 if i = j).

(b) We have △Mji = SF Tp − F T

q R and △Nji = SKTp − KT

q R, sowe analogously make Mji = 0 and Nji = ZT . Since the size of ZT is(q − 1) × p and p > q, by (15) we have ZT = 0.

5 Deformations of contragredient matrix pen-

cils

The canonical form problem for pairs of matrices A ∈ Fm×n, B ∈ Fn×m

under transformations of contragredient equivalence

(A, B) 7→ (SAR−1, RBS−1), S ∈ GL(m, F), R ∈ GL(n, F),

(i.e., for representations of the quiver · ·q

i ) was solved in [2, 3]: each pair isuniquely, up to permutation of cells JF

r (λ) in ⊕iΦF(λi), reduced to a direct

15

sum

(I, C) ⊕

t1⊕

j=1

(Ir1j, Jr1j

) ⊕

t2⊕

j=1

(Jr2j, Ir2j

)

⊕

t3⊕

j=1

(Fr3j, Gr3j

) ⊕

t4⊕

j=1

(Gr4j, Fr4j

) (18)

(we use the notation (14) and put Gr := KTr ), where C is a nonsingular

matrix of the form (4) and ri1 > ri2 > . . . > riti .

Theorem 5.1. One of the simplest miniversal F-deformations of the canon-ical pair (18) under contragredient equivalence over F ∈ {C, R} is the directsum of (I, C) (C is a simplest miniversal F-deformation of C under similar-ity, see Theorem 3.1) and

⊕jIr1j0 0

0 ⊕jJr2j+ H H

0 HP3 H0 Q4

,

⊕jJr1j+ H H H

H ⊕jIr2j0

H 0Q3 0H P4

,

where

Pl =

Frl1+ H H · · · H

Frl2+ H

. . ....

. . . H

0 Frltl+ H

, Ql =

Grl10

H Grl2

.... . .

. . .

H · · · H Grltl

(l = 3, 4), H and H are matrices of the form (6) and (7), the stars denoteindependent parameters.

Proof. Let (A, B) be the canonical matrix pair (18) and let (A,B) be itsdeformation from Theorem 5.1. By Theorem 2.1, we must prove that forevery M ∈ Fm×n, N ∈ Fn×m there exist S ∈ Fm×m and R ∈ Fn×n such that

(M, N) + (SA − AR, RB − BS) = (P, Q),

or, in the block form,

(Mij , Nij) + (SijAj − AiRij , RijBj − BiSij) = (Pij , Qij),

16

where (P, Q) is obtained from (A,B) − (A, B) by replacing its stars withcomplex numbers and is uniquely determined by (M, N).

Therefore, for every pair of summands Pi = (Ai, Bi) and Pj = (Aj , Bj),i 6 j, from the decomposition (18), we must prove that

(a) the pair (Mij , Nij) can be reduced to the pair (Pij , Qij) by transfor-mations (Mij , Nij) 7→ (Mij , Nij) + (△Mij ,△Nij), where

△Mij = SAj − AiR, △Nij = RBj − BiS

with arbitrary R and S; moreover, (Pij, Qij) is uniquely determined (more ex-actly, its entries on the places of stars are uniquely determined) by (Mij , Nij);and, if i < j,

(b) the pair (Mji, Nji) can be reduced to the pair (Pji, Qji) by transfor-mations (Mji, Nji) 7→ (Mji, Nji) + (△Mji,△Nji), where

△Mji = SAi − AjR, △Nji = RBi − BjS

with arbitrary R and S; moreover, (Pji, Qji) is uniquely determined by(Mji, Nji).

Case 1: Pi = (I, JF

p (λ)) and Pj = (I, JF

q (µ)).

(a) We have △Mij = S − R. Make Mij = 0, then S = R;△Nij = RJF

q (µ) − JFp (λ)S. Using Lemma 3.1, we make Nij = 0 if

λ 6= µ, and Nij = H (see (7)) if λ = µ.

(b) We have △Mji = S −R and △Nji = RJF

p (λ)−JF

q (µ)S. As inCase 1(a), make Mji = 0, then Nji = 0 if λ 6= µ and Nji = H if λ = µ.

Case 2: Pi = (I, JF

p (λ)) and Pj = (Jq, I).

(a) We have △Mij = SJq − R. Make Mij = 0, then R = SJq,△Nij = R − JF

p (λ)S = SJq − JF

p (λ)S. Using Lemma 3.1, we makeNij = 0 if λ 6= 0 and Nij = H if λ = 0.

(b) We have △Mji = S − JqR. Make Mji = 0, then S = JqR,△Nji = RJF

p (λ) − S = RJF

p (λ) − JqR. We make Nji = 0 if λ 6= 0 andNji = H if λ = 0.

Case 3: Pi = (I, JF

p (λ)) and Pj = (Fq, Gq).

(a) We have △Mij = SFq − R = S≺ − R. Make Mij = 0, thenR = S≺, △Nij = RGq−JF

p (λ)S = S≺⊳−JF

p (λ)S = SJq−JFp (λ)S. Using

Lemma 3.1, we make Nij = 0 if λ 6= 0 and Nij = H if λ = 0.

17

(b) We have △Mji = S − FqR = S − R▽. Make Mji = 0, thenS = R▽, △Nji = RJF

p (λ) − GqS = RJF

p (λ) − Rg

▽= RJF

p (λ) − Jq−1R.We make Nji = 0 if λ 6= 0 and Nji = H if λ = 0.

Case 4: Pi = (I, JF

p (λ)) and Pj = (Gq, Fq).

(a) We have △Mij = SGq − R. Make Mij = 0, then R = S⊳,△Nij = RFq − JF

p (λ)S = S≺⊳− JF

p (λ)S = SJq−1 − JF

p (λ)S. UsingLemma 3.1, we make Nij = 0 if λ 6= 0 and Nij = H if λ = 0.

(b) We have △Mji = S − GqR = S − Rg. Make Mji = 0, thenS = Rg, △Nji = RJF

p (λ) − FqS = RJF

p (λ) − Rg

▽= RJF

p (λ) − JqR. Wemake Nji = 0 if λ 6= 0 and Nji = H if λ = 0.

Case 5: Pi = (Jp, I) and Pj = (Jq, I). Interchanging the matrices in eachpair, we reduce this case to Case 1.

Case 6: Pi = (Jp, I) and Pj = (Fq, Gq). Interchanging the matrices in eachpair, we reduce this case to Case 4.

Case 7: Pi = (Jp, I) and Pj = (Gq, Fq). Interchanging the matrices in eachpair, we reduce this case to Case 3.

Case 8: Pi = (Fp, Gp) and Pj = (Fq, Gq), i 6 j (and hence p > q).

(a) We have △Nij = RGq −GpS = R⊳−Sg. Make Nij = 0, thenR⊳ = Sg. Further, △Mij = SFq − FpR = S≺ − R▽, so (△Mij)

g =(Sg)≺ − Rg

▽= R≺

⊳− Rg

▽= RJq−1 − Jp−1R and the first row of △Mij

is arbitrary (due to the first row of S). We make the first row ofMij equaling zero. Following the proof of Lemma 3.1 and taking intoaccount that p > q, we make all entries of the (p − 1)× (q − 1) matrixMg

ij equaling zero except for the last row and obtain Mij = H .

(b) We have i < j, △Mji = SFp−FqR = S≺−R▽. Make Mji = 0,then S≺ = R▽. Further, △Nji = RGp − GqS = R⊳ − Sg, (△Nji)

≺ =R≺

⊳−Rg

▽= RJp−1−Jq−1R and the last column of △Nji is arbitrary (due

to the last column of S). We make the last column of △Nji equalingzero. By Lemma 3.1 and the inequality p > q, we make all entries ofthe (q−1)×(p−1) matrix N≺ji equaling zero except for the first columnand obtain Nji = H .

Case 9: Pi = (Fp, Gp) and Pj = (Gq, Fq).

18

(a) We have △Nij = RFq − GpS = R≺ − Sg. Make Nij = 0,then R≺ = Sg, i.e. R = Xg and S = X≺ for an arbitrary X. Further,△Mij = SGq − FpR = S⊳ − R▽ = X≺

⊳− Xg

▽, we make Mij = H .

(b) We have △Mji = SFp − GqR and △Nji = RGp − FqS. So weanalogously make Mji = 0 and Nji = H .

Case 10: Pi = (Gp, Fp) and Pj = (Gq, Fq), i 6 j. Interchanging the matri-ces in each pair, we reduce this case to Case 8.

References

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[2] N. M. Dobrovol’skaya and V. A. Ponomarev, A pair of counter operators,Uspehi Mat. Nauk 20 (no. 6) (1965) 80-86.

[3] R. A. Horn, D. I. Merino, Contragredient equivalence: a canonical formand some applications Linear Algebra Appl. 214 (1995) 43-92.

[4] V. I. Arnold, Geometrical methods in the theory of ordinary differentialequations. Springer-Verlag, New York, 1988.

[5] D. M. Galin, On real matrices depending on parameters, Uspehi Mat.Nauk 27 (no. 1) (1972) 241-242.

[6] A. Edelman, E. Elmroth, B. Kagstrom, A geometric approach to per-turbation theory of matrices and matrix pencils. Part I: Versal deforma-tions, Siam J. Matrix Anal. Appl. 18 (no. 3) (1997) 653-692.

[7] J. Berg, H. Kwantny, A canonical parametrization of the Kronecker formof a matrix pencil, Automatica 31 (1995) 669-680.

[8] I. Garcıa-Planas, D. Magret, Deformation and stability of triples of ma-trices, Linear Algebra Appl. 254 (1997) 159-192.

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